proof of Abel lemma (by expansion)
1 Abel lemma
$$\sum _{i=0}^{n}{a}_{i}{b}_{i}=\sum _{i=0}^{n1}{A}_{i}({b}_{i}{b}_{i+1})+{A}_{n}{b}_{n},$$  (1) 
where, ${A}_{i}={\sum}_{k=0}^{i}{a}_{k}$. Sequences $\{{a}_{i}\}$, $\{{b}_{i}\}$, $i=0,\mathrm{\dots},n$, are real or complex one.
2 Proof
We consider the expansion of the sum
$$\sum _{i=0}^{n}{A}_{i}({b}_{i}{b}_{i+1})$$ 
on two different forms, namely:

1.
On the short way.
$$\sum _{i=0}^{n}{A}_{i}({b}_{i}{b}_{i+1})=\sum _{i=0}^{n1}{A}_{i}({b}_{i}{b}_{i+1})+{A}_{n}{b}_{n}{A}_{n}{b}_{n+1}.$$ (2) 
2.
On the long way.
$$\sum _{i=0}^{n}{A}_{i}({b}_{i}{b}_{i+1})=\sum _{i=0}^{n}{A}_{i}{b}_{i}\sum _{i=0}^{n}{A}_{i}{b}_{i+1}=\sum _{i=0}^{n}{A}_{i}{b}_{i}\sum _{i=1}^{n+1}{A}_{i1}{b}_{i}=$$ 
$${A}_{0}{b}_{0}+\sum _{i=1}^{n}({A}_{i1}+{a}_{i}){b}_{i}\sum _{i=1}^{n}{A}_{i1}{b}_{i}{A}_{n}{b}_{n+1}=\sum _{i=0}^{n}{a}_{i}{b}_{i}{A}_{n}{b}_{n+1},$$  (3) 
where a simplification has been performed. Notice that ${A}_{0}={a}_{0}$. By equating (2), (3), the last two terms cancel, ^{1}^{1}Without loss of generality, ${b}_{n+1}$ may be assumed finite. Indeed we don’t need ${b}_{n+1}$, but the proof is a couple lines larger. It is left as an exercise. and then, (1) follows. $\mathrm{\square}$
Title  proof of Abel lemma (by expansion) 

Canonical name  ProofOfAbelLemmabyExpansion 
Date of creation  20130322 17:28:14 
Last modified on  20130322 17:28:14 
Owner  perucho (2192) 
Last modified by  perucho (2192) 
Numerical id  7 
Author  perucho (2192) 
Entry type  Proof 
Classification  msc 40A05 